Electronic – Charging a capacitor parallel to a resistor

It's not clear what exactly you mean by series versus parallel resistor-capacitor filters. Posting a schematic of each would clarify this.

You probably mean series is something the signal passes thru, and parallel is something that works as a shunt. Note that the same thing that is a low pass filter in series is a high pass filter as a shunt, and vice versa.

Basically a capacitor will block low frequencies and short high ones. If you put it in series with a signal then it is a high pass filter. If you put it accross a signal, it will short the high frequencies thereby making a low pass filter. The value of the capacitance and the resistance it is working against tell you the -3dB rolloff frequency of the filter, whether high pass or low pass. This frequency is:

\$ f = \dfrac{1}{2 \cdot \pi \cdot R \cdot C} \$

When R is in Ohms and C in Farads, then f is in Hertz.

The basic formula for a capacitor is Q = CV. If this is differentiated you get: -

\$\dfrac{dq}{dt} = C\dfrac{dv}{dt}\$ and this equals current.

So, for a given current and a given capacitance the voltage rises at a rate of I/C.

You appear to be saying that the circuit that generates the current is "fixed" and so the only option I can see is make the capacitance "several" times bigger in order to lengthen the charge time.

Putting a resistor in parallel will not get you what you want because the capacitor will not charge up to the full voltage.

A more complicated solution is to create a constant current sink (and this is not a simple resistor as per your shunt resistor idea). The sink would be in parallel with the capacitor and basically this diverts current away from the capacitor making the net current into the capacitor smaller and hence increase the charge time.